# Trajectory

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Trajectory

Trajectory is the path a moving object follows through space. The object might be a projectile or a satellite, for example. It thus includes the meaning of orbit - the path of a planet, an asteroid or a comet as it travels around a central mass. A trajectory can be described mathematically either by the geometry of the path, or as the position of the object over time.

In control theory a trajectory is a time-ordered set of states of a dynamical system (see e.g. Poincaré map). In discrete mathematics, a trajectory is a sequence$\left(f^k\left(x\right)\right)_\left\{k in mathbb\left\{N$ of values calculated by the iterated application of a mapping$f$ to an element $x$ of its source.

The word trajectory is also often used metaphorically, for instance, to describe an individual's career.

Physics of trajectories

A familiar example of a trajectory is the path of a projectile such as a thrown ball or rock. In a greatly simplified model the object moves only under the influence of a uniform homogenous gravitational force field. This can be a good approximation for a rock that is thrown for short distances for example, at the surface of the moon. In this simple approximation the trajectory takes the shape of a parabola. Generally, when determining trajectories it may be necessary to account for nonuniform gravitational forces, air resistance (drag and aerodynamics). This is the focus of the discipline of ballistics.

One of the remarkable achievements of Newtonian mechanics was the derivation of the laws of Kepler, in the case of the gravitational field of a single point mass (representing the Sun). The trajectory is a conic section, like an ellipse or a parabola. This agrees with the observed orbits of planets and comets, to a reasonably good approximation. Although if a comet passes close to the Sun, then it is also influenced by other forces, such as the solar wind and radiation pressure, which modify the orbit, and cause the comet to eject material into space.

Newton's theory later developed into the branch of theoretical physics known as classical mechanics. It employs the mathematics of differential calculus (which was, in fact, also initiated by Newton, in his youth). Over the centuries, countless scientists contributed to the development of these two disciplines. Classical mechanics became a most prominent demonstration of the power of rational thought, i.e. reason, in science as well as technology. It helps to understand and predict an enormous range of phenomena. Trajectories are but one example.

Consider a particle of mass $m$, moving in a potential field $V$. Physically speaking, mass represents inertia, and the field $V$ represents external forces, of a particular kind known as "conservative". That is, given $V$ at every relevant position, there is a way to infer the associated force that would act at that position, say from gravity. Not all forces can be expressed in this way, however.

The motion of the particle is described by the second-order differential equation

:$m frac\left\{mathrm\left\{d\right\}^2 vec\left\{x\right\}\left(t\right)\right\}\left\{mathrm\left\{d\right\}t^2\right\} = - abla V\left(vec\left\{x\right\}\left(t\right)\right)$ with $vec\left\{x\right\} = \left(x, y, z\right)$

On the right-hand side, the force is given in terms of $abla V$, the gradient of the potential, taken at positions along the trajectory. This is the mathematical form of Newton's second law of motion: mass times acceleration equals force, for such situations.

Examples

Uniform gravity, no drag or wind

The case of uniform gravity, disregarding drag and wind, yields a trajectory which is a parabola. To model this, one chooses $V = m g z$, where $g$ is the acceleration of gravity. This gives the equations of motion

:$frac\left\{mathrm\left\{d\right\}^2 x\right\}\left\{mathrm\left\{d\right\}t^2\right\} = frac\left\{mathrm\left\{d\right\}^2 y\right\}\left\{mathrm\left\{d\right\}t^2\right\} = 0$ :$frac\left\{mathrm\left\{d\right\}^2 z\right\}\left\{mathrm\left\{d\right\}t^2\right\} = - g$

Simplifications are made for the sake of studying the basics. The actual situation, at least on the surface of Earth, is considerably more complicated than this example would suggest, when it comes to computing actual trajectories. By deliberately introducing such simplifications, into the study of the given situation, one does, in fact, approach the problem in a way that has proved exceedingly useful in physics.

The present example is one of those originally investigated by Galileo Galilei. To neglect the action of the atmosphere, in shaping a trajectory, would (at best) have been considered a futile hypothesis by practical minded investigators, all through the Middle Ages in Europe. Nevertheless, by anticipating the existence of the vacuum, later to be demonstrated on Earth by his collaborator Evangelista Torricelli, Galileo was able to initiate the future science of mechanics. And in a near vacuum, as it turns out for instance on the Moon, his simplified parabolic trajectory proves essentially correct.

Relative to a flat terrain, let the initial horizontal speed be $v_h,$, and the initial vertical speed be $v_v,$. It will be shown that, the range is $2v_h v_v/g,$, and the maximum altitude is $\left\{v_v^2\right\}/2g,$. The maximum range, for a given total initial speed $v$, is obtained when $v_h=v_v,$, i.e. the initial angle is 45 degrees. This range is $v^2/g,$, and the maximum altitude at the maximum range is a quarter of that.

Derivation

The equations of motion may be used to calculate the characteristics of the trajectory.

Let :$p\left(t\right);$ be the position of the projectile, expressed as a vector:$t;$ be the time into the flight of the projectile,:$v_h ;$ be the initial horizontal velocity (which is constant):$v_v ;$ be the initial vertical velocity upwards.The path of the projectile is known to be a parabola so:$p\left(t\right) = \left( A t, 0 , a t^2 + b t + c \right),$where $A,,a,,b,,c$ are parameters to be found. The first and second derivatives of $p$ are::$p\text{'}\left(t\right) = \left( A , 0 , 2 a t + b \right),quad p"\left(t\right) = \left( 0 , 0 , 2 a \right).$At $t=0$ :$p\left(0\right)= \left(0, 0, 0\right) p\text{'}\left(0\right)=\left(v_h,0,v_v\right), p"\left(0\right)=\left(0,0,-g\right)$so:$A = v_h, a = -g/2, b = v_v, c = 0$. This yields the formula for a parabolic trajectory::$p\left(t\right) = \left(v_h t,0,v_v t - g t^2/2\right),qquad$ (Equation I: trajectory of parabola).

Range and height

The range $R$ of the projectile is found when the $z$-component of $p$ is zero, that is when:$0 = v_v t - g t^2/2 = t left\left( v_v - g t/2 ight\right),$which has solutions at $t=0$ and$t = 2 v_v /g$(the hang-time of the projectile).The range is then$R = 2 v_h v_v/g.,$

From the symmetry of the parabola the maximum height occurs at the halfway point $t=v_v/g$ at position:$p\left(v_v/g\right)=\left(v_h v_v/g,0,v_v^2/\left(2g\right)\right),$This can also be derived by finding when the $z$-component of $p\text{'}$ is zero.

Angle of elevation

In terms of angle of elevation $heta$ and initial speed $v$::$v_h=v cos heta,quad v_v=v sin heta ;$giving the range as:$R= 2 v^2 cos\left( heta\right) sin\left( heta\right) / g = v^2 sin\left(2 heta\right) / g,.$This equation can be rearranged to find the angle for a required range:$\left\{ heta \right\} = frac 1 2 sin^\left\{-1\right\} left\left( \left\{ \left\{g R\right\} over \left\{ v^2 \right\} \right\} ight\right)$ (Equation II: angle of projectile launch)Note that the sine function is such that there are two solutions for $heta$ for a given range $d_h$. Physically, this corresponds to a direct shot versus a mortar shot up and over obstacles to the target.The angle $heta$ giving the maximum range can be found by considering the derivative or $R$ with respect to $heta$ and setting it to zero.:$\left\{mathrm\left\{d\right\}Rover mathrm\left\{d\right\} heta\right\}=\left\{2v^2over g\right\} cos\left(2 heta\right)=0$which has a non trivial solutions at $2 heta=pi/2=90^circ$.The maximum range is then $R_\left\{max\right\} = v^2/g,$. At this angle $sin\left(pi/2\right)=1$ so the maximum height obtained is $\left\{v^2 over 4g\right\}$.

To find the angle giving the maximum height for a given speed calculate the derivative of the maximum height $H=v^2 sin\left( heta\right) /\left(2g\right)$ with respect to $heta$, that is $\left\{mathrm\left\{d\right\}Hover mathrm\left\{d\right\} heta\right\}=v^2 cos\left( heta\right) /\left(2g\right)$which is zero when $heta=pi=180^circ$. So the maximum height $H_\left\{max\right\}=\left\{v^2over 2g\right\}$ is obtain when the projectile is fired straight up.The equation of the trajectory of a projectile fired in uniform gravity in a vacuum on Earth in Cartesian coordinates is

$y=-\left\{gsec^2 hetaover 2v_0^2\right\}x^2+x an heta+h$,

where "v"0 is the initial speed, "h" is the height the projectile is fired from, and "g" is the acceleration due to gravity).

Uphill/downhill in uniform gravity in a vacuum

Given a hill angle $alpha$ and launch angle $heta$ as before, it can be shown that the range along the hill $R_s$ forms a ratio with the original range $R$ along the imaginary horizontal, such that::$frac\left\{R_s\right\} \left\{R\right\}=\left(1-cot heta an alpha\right)sec alpha$ (Equation 11)

In this equation, downhill occurs when $alpha$ is between 0 and -90 degrees. For this range of $alpha$ we know: $an\left(-alpha\right)=- an alpha$ and $sec \left( - alpha \right) = sec alpha$. Thus for this range of $alpha$,$R_s/R=\left(1+ an heta an alpha\right)sec alpha$. Thus $R_s/R$ is a positive value meaning the range downhill is always further than along level terrain. The lower level of terrain causes the projectile to remain in the air longer, allowing it to travel further horizontally before hitting the ground.

While the same equation applies to projectiles fired uphill, the interpretation is more complex as sometimes the uphill range may be shorter or longer than the equivalent range along level terrain. Equation 11 may be set to $R_s/R=1$ (i.e. the slant range is equal to the level terrain range) and solving for the "critical angle" $heta_\left\{cr\right\}$::$1=\left(1- an heta an alpha\right)sec alpha quad ;$ :$heta_\left\{cr\right\}=arctan\left(\left(1-csc alpha\right)cot alpha\right) quad ;$

Equation 11 may also be used to develop the "rifleman's rule" for small values of $alpha$ and $heta$ (i.e. close to horizontal firing, which is the case for many firearm situations). For small values, both $an alpha$ and $an heta$ have a small value and thus when multiplied together (as in equation 11), the result is almost zero. Thus equation 11 may be approximated as::$frac\left\{R_s\right\} \left\{R\right\}=\left(1-0\right)sec alpha$And solving for level terrain range, $R$:$R=R_s cos alpha$ "Rifleman's rule"Thus if the shooter attempts to hit the level distance R, s/he will actually hit the slant target. "In other words, pretend that the inclined target is at a horizontal distance equal to the slant range distance multiplied by the cosine of the inclination angle, and aim as if the target were really at that horizontal position." [http://www.snipertools.com/article4.htm]

Derivation based on equations of a parabola

The intersect of the projectile trajectory with a hill may most easily be derived using the trajectory in parabolic form in Cartesian coordinates (Equation 10) intersecting the hill of slope $m$ in standard linear form at coordinates $\left(x,y\right)$::$y=mx+b ;$ (Equation 12) where in this case, $y=d_v$, $x=d_h$ and $b=0$

Substituting the value of $d_v=m d_h$ into Equation 10::$m x=-frac\left\{g\right\}\left\{2v^2\left\{cos\right\}^2 heta\right\}x^2 + frac\left\{sin heta\right\}\left\{cos heta\right\} x$:$x=frac\left\{2v^2cos^2 heta\right\}\left\{g\right\}left\left(frac\left\{sin heta\right\}\left\{cos heta\right\}-m ight\right)$ (Solving above x)This value of x may be substituted back into the linear equation 12 to get the corresponding y coordinate at the intercept::$y=mx=m frac\left\{2v^2cos^2 heta\right\}\left\{g\right\} left\left(frac\left\{sin heta\right\}\left\{cos heta\right\}-m ight\right)$Now the slant range $R_s$ is the distance of the intercept from the origin, which is just the hypotenuse of x and y::$R_s=sqrt\left\{x^2+y^2\right\}=sqrt\left\{left\left(frac\left\{2v^2cos^2 heta\right\}\left\{g\right\}left\left(frac\left\{sin heta\right\}\left\{cos heta\right\}-m ight\right) ight\right)^2+left\left(m frac\left\{2v^2cos^2 heta\right\}\left\{g\right\} left\left(frac\left\{sin heta\right\}\left\{cos heta\right\}-m ight\right) ight\right)^2\right\}$::$=frac\left\{2v^2cos^2 heta\right\}\left\{g\right\} sqrt\left\{left\left(frac\left\{sin heta\right\}\left\{cos heta\right\}-m ight\right)^2+m^2 left\left(frac\left\{sin heta\right\}\left\{cos heta\right\}-m ight\right)^2\right\}$::$=frac\left\{2v^2cos^2 heta\right\}\left\{g\right\} left\left(frac\left\{sin heta\right\}\left\{cos heta\right\}-m ight\right) sqrt\left\{1+m^2\right\}$

Now $alpha$ is defined as the angle of the hill, so by definition of tangent, $m= an alpha$. This can be substituted into the equation for $R_s$::$R_s=frac\left\{2v^2cos^2 heta\right\}\left\{g\right\} left\left(frac\left\{sin heta\right\}\left\{cos heta\right\}- an alpha ight\right) sqrt\left\{1+ an^2 alpha\right\}$Now this can be refactored and the trigonometric identity for $sec alpha = sqrt \left\{1 + an^2 alpha\right\}$ may be used::$R_s=frac\left\{2v^2cos hetasin heta\right\}\left\{g\right\}left\left(1-frac\left\{sin heta\right\}\left\{cos heta\right\} analpha ight\right)secalpha$Now the flat range $R=v^2sin 2 heta / g = 2v^2sin hetacos heta / g$ by the previously used trigonometric identity and $sin heta/cos heta=tan heta$ so::$R_s=R\left(1- an heta analpha\right)secalpha ;$:$frac\left\{R_s\right\}\left\{R\right\}=\left(1- an heta analpha\right)secalpha$

Orbiting objects

If instead of a uniform downwards gravitational force we considertwo bodies orbiting with the mutual gravitation between them, we obtain
Kepler's laws of planetary motion. The derivation of these was one of the major works of Isaac Newton and provided much of the motivation for the development of differential calculus.

ee also

*Aft-crossing trajectory
*Orbit (dynamics)
*Orbit (group theory)
*Planetary orbit
*Porkchop plot
*Rigid body
*Trajectory of a projectile

* [http://www.physics-lab.net/applets/projectile-motion Projectile Motion Flash Applet]
* [http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html Trajectory calculator]
* [http://www.phy.hk/wiki/englishhtm/ThrowABall.htm An interactive simulation on projectile motion]
* [http://publicliterature.org/tools/projectile_motion/ Projectile Motion Simulator, java applet]
* [http://www.thewritingpot.com/projectilelab/ Projectile Lab, JavaScript trajectory simulator]
* [http://www.excelcalcs.com/content/view/74/109/ Projectile calculation in MS Excel] – calculation of the projectile position after a given time, the maximum height reached and the range of the projectile. The projectile path is plotted on an Excel chart and all cell formulae are shown in mathematical notation.
* [http://demonstrations.wolfram.com/ParabolicProjectileMotionShootingAHarmlessTranquilizerDartAt/ Parabolic Projectile Motion: Shooting a Harmless Tranquilizer Dart at a Falling Monkey] by Roberto Castilla-Meléndez, Roxana Ramírez-Herrera, and José Luis Gómez-Muñoz, The Wolfram Demonstrations Project.
* [http://scienceworld.wolfram.com/physics/Trajectory.html Trajectory] , ScienceWorld.

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### Look at other dictionaries:

• Trajectory — Tra*ject o*ry, n.; pl. {Trajectories}. [Cf. F. trajectoire.] The curve which a body describes in space, as a planet or comet in its orbit, or stone thrown upward obliquely in the air. [1913 Webster] …   The Collaborative International Dictionary of English

• trajectory — (n.) 1690s, from Mod.L. trajectoria, from fem. of trajectorius of or pertaining to throwing across, from L. traiectus thrown over or across, pp. of traicere throw across, from L. trans across (see TRANS (Cf. trans )) + icere, combining form of… …   Etymology dictionary

• trajectory — [n] course curve, direction, flight, flow, line, movement, orbit, path, range, route, track, trail; concepts 501,514 …   New thesaurus

• trajectory — ► NOUN (pl. trajectories) ▪ the path described by a projectile flying or an object moving under the action of given forces. ORIGIN Latin trajectoria, from traicere throw across …   English terms dictionary

• trajectory — [trə jek′tə rē] n. pl. trajectories [ML trajectorius < L trajectus: see TRAJECT] 1. the curved path of something hurtling through space, esp. that of a projectile from the time it leaves the muzzle of the gun 2. Math. a) a curve or surface… …   English World dictionary

• trajectory — [[t]trəʤe̱ktəri[/t]] trajectories 1) N COUNT: with supp The trajectory of a moving object is the path that it follows as it moves. ...the trajectory of an artillery shell. 2) N COUNT: with supp The trajectory of something such as a person s… …   English dictionary

• trajectory — UK [trəˈdʒekt(ə)rɪ] / US [trəˈdʒektərɪ] noun [countable] Word forms trajectory : singular trajectory plural trajectories 1) the high curving line in which an object such as a missile moves through the air The bomb followed a high trajectory… …   English dictionary

• trajectory — noun (plural ries) Etymology: New Latin trajectoria, from feminine of trajectorius of passing, from Latin traicere to cause to cross, cross, from trans , tra trans + jacere to throw more at jet Date: 1696 1. the curve that a body (as a planet or… …   New Collegiate Dictionary

• trajectory — tra|jec|to|ry [ trə dʒektəri ] noun count 1. ) TECHNICAL the high curving line in which an object such as a missile moves through the air: The bomb followed a high trajectory toward its target. 2. ) FORMAL the way in which a process or event… …   Usage of the words and phrases in modern English

• trajectory — See ballistic trajectory …   Military dictionary